parabolic isometries of hyperbolic spaces and...

Parabolic isometries of hyperbolic spacesand discreteness

John R ParkerDurham University, UK

[email protected]

Shige

It is a great privilege for me to speak at the conference in honourof Shigeyasu Kamiya on the occasion of his retirement.

I have known Shige for many years.

These are pictures of us

I At the 2007 conference for Alan Beardon’s retirement

I In my home in Durham

I At the conference Shige organised in Okayama in 1998

Shimizu’s lemma

Lemma (Shimizu 1963)

Let T and S be the following matrices in SL(2,R)

T =

(1 t0 1

)S =

(a bc d

).

Suppose c 6= 0. If the group Γ = 〈T , S〉 is discrete then |ct| ≥ 1.

The proof has three steps.Step 1Consider the sequence in Γ defined by S0 = S and Sj+1 = SjTS

−1j .

Write Sj =

(aj bjcj dj

). Then Sj+1 is given by(

aj bjcj dj

)(1 t0 1

)(dj −bj−cj aj

)=

(1− ajcj t a2

j t

−c2j t 1 + ajcj t

).

In particular cj+1 = −c2j t

Proof of Shimizu’s lemma continued

Step 2From the sequence {Sj} construct a dynamical system:

|cj+1t| = |cj t|2 and so |cj t| = |c0t|2j

= |ct|2j .Find a condition that ensures we lie in a finite basin of attraction:If |ct| = |c0t| < 1 then |cj t|, and hence cj , tends to 0.Moreover, since c 6= 0 then cj 6= 0.

Step 3We use cj t −→ 0 to show Sj tends to T :We have aj+1 − 1 = −ajcj t = −(aj − 1)cj t − cj t, so for j ≥ 1:

|aj − 1| ≤ |a0 − 1| |c0t|2j

+ j |c0t|2j

Hence aj tends to 1.

Using our expression for Sj+1, this shows Sj tends to T .

Since cj 6= 0 then Sj 6= T and Γ = 〈T ,S〉 is not discrete.

Hence if Γ is discrete, we must have |ct| ≥ 1. �

Some hyperbolic geometry

I A matrix S in SL(2,R) acts on the upper half planeas a Mobius transformation S(z) in PSL(2,R)

S =

(a bc d

)corresponds to S(z) = az+b

cz+d .

I S(z) is an isometry of the hyperbolic plane H2 = H2R

I A discrete subgroup Γ of SL(2,R) actsproperly discontinuously on H2

R.

I The quotient M = H2R/Γ is an orbifold.

I The matrix T corresponds to the Mobius transformationT (z) = z + t.This has (Euclidean) translation length `T = |t|

I For u > 0 the horoball Hu of height u at ∞ isHu = {z = x + iy ∈ H2

R : y > u}I A horoball at a point x ∈ R is an open disc tangent to R at x

Isometric spheres

Let S(z) = (az + b)/(cz + d) ∈ PSL(2,R) not fixing ∞So c 6= 0. The isometric sphere I (S) of S is the Euclideansemi-circle with centre S−1(∞) = −d/c and radius rs = 1/|c|

−d/c=S ( ) a/c=S( )−1

I(S) −1

8 8

1/| |cI(S )

u

1/ | |u c 2

Hu

HS( )u

S sends the outside of I (S) to the inside of I (S−1).

S sends the horoball Hu of height u centred at ∞to a horoball S(Hu) of diameter 1/u|c |2 at S(∞).

So if u ≥ rS = 1/|c | then Hu and S(Hu) are disjoint.

Geometric interpretation of Shimizu’s lemma

Let Γ be a discrete subgroup of PSL(2,R) containingT (z) = z + t where t > 0.

(1) If S is any element of Γ not fixing ∞then the radius rS of the isometric sphere of S satisfiesrS ≤ `T(Note this is an inequality between two Euclidean quantities)Proof: rs = 1/|c |, `T = |t| and Shimizu says |ct| ≥ 1.

(2) Horoball Ht of height u = t is precisely invariant under ΓThat is, for any S ∈ Γ either S(Ht) = Ht or S(Ht) ∩ Ht = ∅.

On the orbifold M = H2R/Γ

C = Ht/Γ∞ is a cusp neighbourhood of hyperbolic area 1.

C

M

The modular group

Shimizu’s lemma is sharp for the modular group PSL(2,Z)In this case, t = 1 and S(z) = −1/z has rS = 1.There is a cusp neighbourhood C which cannot be enlarged.It has area is 1, which is large compared to thearea of H2

R/PSL(2,Z), which is π/3

0 1−1 2

C

Generalisations of Shimizu’s lemma

I Leutbecher 1967 Subgroups of PSL(2,C) = Isom0(H3R)

containing a translation.

I Wielenberg 1977 Subgroups of PO0(n, 1) = Isom0(HnR)

containing a translation.

I Kamiya 1983 Subgroups of PU(n, 1) = Isom0(HnC) or

PSp(n, 1) = Isom0(HnH) containing a vertical Heisenberg

translation.

I Apanasov 1985, Ohtake 1985 Subgroups of Isom(HnR) with

n ≥ 4 containing a screw parabolic map: there is no uniformbound on radii of isometric spheres.

I JRP 1992 Subgroups of PU(n, 1) containing a non-verticalHeisenberg translation: there is no uniform bound on radii ofisometric spheres.

I Waterman 1993 Subgroups of Isom(HnR) with n ≥ 4

containing a screw parabolic map: radii of isometric spheresbounded in terms of parabolic translation length at centres.

Jørgensen’s inequalityJørgensen 1976 If 〈T , S〉 subgroup of SL(2,C) discrete, thenelementary or |tr2(S)− 4|+ |tr[S ,T ]− 2| ≥ 1(Same structure of proof as Shimizu’s lemma).

While working on basin of attraction part of this problemBrooks & Matelski 1978/1979 produced

Possibly the first picture of the Mandelbrot set

Shimizu’s lemma for real hyperbolic space

A parabolic isometry T of HnR fixing ∞ acts on Rn−1 as

T (x) = Ux + t where U ∈ O(n − 1), t ∈ Rn−1 and Ut = t.

Let NU = max{‖(U − I )x‖ : ‖x‖ = 1}The Euclidean translation length of T at x is`T (x) = ‖T (x)− x‖ = ‖(U − I )x + t‖ =

√‖(U − I )x‖2 + ‖t‖2

Theorem (Waterman 1993)

Let Γ < Isom(HnR) be discrete and contain T (x) = Ux + t.

Suppose NU < 1/4 and write K = 12 (1 +

√1− 4NU).

Let S ∈ Γ not fixing ∞ have isometric sphere of radius rS .

Then r2S ≤

`T (S−1(∞))`T (S(∞))

K 2.

I If U = I then NU = 0 and K = 1.Also, `T (x) = ‖t‖ and Waterman gives rS ≤ ‖t‖This is Wielenberg’s version of Shimizu’s lemma.

We want to generalise this to other hyperbolic spaces

Hyperbolic spaces

Let F be one of

I the real numbers R,

I the complex numbers C,

I the quaternions HLet Fn,1 be the n + 1 dimensional F-vector space(with scalars in F acting on the right) equipped with〈·, ·〉 Hermitian form (bilinear for Rn,1) of signature (n, 1)Let V− = {z ∈ Fn,1 : 〈z, z〉 < 0}and V0 = {z ∈ Fn,1 − {0} : 〈z, z〉 = 0}Let P : Fn,1 − {0} −→ FPn be the (right) projection map.

Then HnF = PV− and ∂Hn

F = PV0; metric on HnF is given by:

ds2 =−4

〈z, z〉2det

(〈z, z〉〈dz, z〉〈z, dz〉〈dz, dz〉

)The hyperbolic spaces are Hn

R, HnC, Hn

H together with H2O

where O are the octonions (see Chen & Greenberg 1974).

More about hyperbolic spaces

Let O(n, 1), U(n, 1), Sp(n, 1) be the group preserving 〈·, ·〉when F = R, C, H respectively (acting on Fn,1 on the left).This group acts (projectively) by isometries on Hn

F(For F = O there is no vector space O2,1

but there is an analogous isometry group F4(−20) of H2O.

We will not consider this case here.)We pass between matrix groups and isometries without comment.

An isometry S of a hyperbolic space H is

I loxodromic (or hyperbolic) if it has two fixed points, both on∂H

I parabolic if it has a unique fixed point, lying on ∂H

I elliptic if it fixes (at least) one point of H

There are finer classifications of these types.We will mainly be interested in parabolic maps, which are:Either (Heisenberg)-translations or screw parabolic maps.

Shimizu’s lemma for other hyperbolic spaces

I Kamiya 1983 Subgroups of SU(n, 1) or Sp(n, 1) containing avertical Heisenberg translation.

I Hersonsky-Paulin 1996 Subgroups of SU(n, 1) containing anon-vertical Heisenberg translation (not given geometrically).

I JRP 1997 Subgroups of SU(n, 1) containing a non-verticalHeisenberg translation.

I I. Kim & JRP 2003 Subgroups of Sp(n, 1) containing anon-vertical Heisenberg translation.

I Jiang & JRP 2003 Subgroups of SU(2, 1) containing a screwparabolic map (not given geometrically).

I D. Kim 2004 Subgroups of Sp(2, 1) containing (certain typesof) screw parabolic maps (not given geometrically).

I Kamiya & JRP 2008 Subgroups of SU(2, 1) containing apositively oriented screw parabolic map.

I Cao & JRP 2014 Subgroups of SU(n, 1) or Sp(n, 1)containing any parabolic map.

Other generalisations

The stable basin theorem

I Basmajian & Miner 1998 Stable basin theorem – strongerhypothesis than Shimizu/Jørgensen. Includes version ofShimizu’s lemma for SU(2, 1)

I Kamiya 2000, Kamiya & JRP 2002 SBT for Heisenbergtranslations follows from Shimizu’s lemma.

Generalisations of Jørgensen’s inequality:

I Jiang, Kamiya & JRP 2003 Jørgensen’s inequality forsubgroups of SU(2, 1)

I Markham 2003 Jørgensen’s inequality for subgroups ofPSp(2, 1) and (some) subgroups of F4(−20)

I D. Kim 2004 Jørgensen’s inequality for subgroups of PSp(2, 1)

I Cao & JRP 2011 Jørgensen’s inequality for subgroups ofSU(n, 1) or Sp(n, 1).

Heisenberg groups and the boundary of hyperbolic spaces

We can identify

I ∂HnR with Rn−1 ∪ {∞},

I ∂HnC with N2n−1 ∪ {∞},

I ∂HnH with N4n−1 ∪ {∞}.

N2n−1 = Cn−1 × R is the (2n − 1)-dimensional Heisenberg group,and N4n−1 = Hn−1 × R3 = Hn−1 ×=H is the(4n − 1)-dimensional generalised Heisenberg groupboth with the group law(ζ1, v1) · (ζ2, v2) = (ζ1 + ζ2, v1 + v2 + 2=(ζ∗2ζ1))(where z∗ is the conjugate transpose)We will write N for both cases

The Cygan metric on N is the metric associated to the norm

‖(ζ, v)‖ =(‖ζ‖4 + |v |2

)1/4

It generalises the Euclidean metric on Rn−1 for HnR and

the square root of the Euclidean metric on R for H1C ≈ H2

R.

Action of PU(n, 1) and PSp(n, 1) on NWe write an element S of PU(n, 1) or PSp(n, 1) and its inverse as

S =

a γ∗ bα A βc δ∗ d

, S−1 =

d β∗ bδ A∗ γc α∗ a

where a, b, c , d ∈ F, α, β, γ, δ ∈ Fn−1, A ∈ U(n − 1) or Sp(n − 1).If c = 0 then S fixes ∞; if c 6= 0 we define isometric spheres.

The isometric sphere I (S) of S is the Cygan sphere of radiusrS = 1/|c |1/2 with centre S−1(∞) =

(δc−1/

√2,=(dc−1)

)∈ N.

S sends the outside of I (S) to the inside of I (S−1).

Pictures of Cygan spheres and hemispheres by Anton Lukyanenko.

Heisenberg translations in PU(n, 1) and PSp(n, 1)

The simplest parabolic maps in PU(n, 1) and PSp(n, 1) areHeisenberg translations:

The (generalised) Heisenberg group N acts on itself by lefttranslation: T(τ,t) : (ζ, v) 7−→

(ζ + τ, v + t + 2=(ζ∗τ)

)As a matrix in PU(n, 1) or PSp(n, 1) it is

T(τ,t) = T =

1 −√

2τ∗ −‖τ‖2 + t

0 I√

2τ0 0 1

.

Note: t in top right hand entry is pure imaginaryso is it in complex case.

A Heisenberg translation by (0, t) is called a vertical translationand lies in the centre of N.

The Cygan translation length of T at (ζ, v) ∈ N is

`T ((ζ, v)) =(‖τ‖4 + |t + 4=(ζ∗τ)|2

)1/4.

Shimizu’s lemma for Heisenberg translations

Theorem (JRP 1997, Kim-JRP 2003)

Let Γ < PU(n, 1) or PSp(n, 1) be discrete and containHeisenberg translation T by (τ, t).Let S ∈ Γ not fixing ∞ have isometric sphere of radius rS .Then r2

S ≤ `T (S−1(∞))`T (S(∞)) + 4‖τ‖2.

I When τ = 0 then `T ((ζ, v)) = |t|1/2 get r2S ≤ `2

T = |t|(that is |ct| ≥ 1), due to Kamiya 1983.

Structure of proof same as for classical Shimizu:

I Consider the sequence S0 = S , Sj+1 = SjTS−1j

I Show in finite basin of attraction of dynamical systemI Deduce Sj tends to T .

The variables in the dynamical system are:Xj =

(max{`T (S−1

j (∞)), `T (Sj(∞))}/rSj)2, Yj =

(‖τ‖/rSj

)2

They satisfy Xj+1 ≤ X 2j + 4Yj , Yj+1 ≤ XjYj

If bound on rS is not true then we show Xj + 4Yj < 1and Xj , Yj tend to 0 as j →∞

Invariant horoballs

We can give HnF the structure N× R+

A horoball Hu of height u at ∞ is N× (u,∞).

I For vertical translations by (0, t)the horoball H|t| is precisely invariant.

I For non-vertical translations by (τ, t) with τ 6= 0there is a precisely invariant sub-horospherical region.We will not go into details about these.

I There is a sharp version of Shimizu’s lemma for PU(2, 1)yielding a cusp neighbourhood of volume 1/4This cusp neighbourhood is as maximal for

the Eisenstein-Picard lattice PU(2, 1;Z[ 1+i√

32 ]) and its sister

– which have covolume π2/27

Positively oriented screw parabolic maps in PU(2, 1)

Consider T =

1 0 it0 e iθ 00 0 1

∈ PU(2, 1)

T is positively oriented if t sin(θ) > 0.T acts on N3 as T : (ζ, v) 7−→ (e iθζ, v + t)Its Cygan translation length `T ((ζ, v)) at (ζ, v) is:∣∣2|ζ|2(e iθ − 1) + it

∣∣1/2=(|ζ|4|e iθ − 1|4 + (2|ζ|2 sin(θ) + t)2

)1/4

We have the following version of Shimizu’s lemma for groups withpositively oriented screw parabolic maps (cf Waterman’s theorem):

Theorem (Kamiya-JRP 2008)

Let Γ < PU(2, 1) be discrete and contain positively oriented T .Suppose |e iθ − 1| < 1/4 and write K = 1

2 (1 +√

1− 4|e iθ − 1|).Let S ∈ Γ not fixing ∞ have isometric sphere of radius rS .

Then r2S ≤

`T (S−1(∞))`T (S(∞))

K 2.

I Note that if θ = 0 we obtain Kamiya’s 1983 result.

General parabolic maps in PU(n, 1)

I A general parabolic map in PU(n, 1) has the form

T =

1 −√

2τ∗ −‖τ‖2 + it

0 U√

2τ0 0 1

where U ∈ U(n − 1) with Uτ = τ(so if n = 2 and U 6= I then τ = 0).

I If U 6= I then this is a screw parabolic map.

I T acts on N2n−1 asT : (ζ, v) 7−→ (Uζ + τ, v + t + 2=(ζ∗τ))

I Its Cygan translation length at (ζ, v) is

`T ((ζ, v)) =(‖(U−I )ζ+τ‖4+|t+2=((ζ∗−τ∗)(Uζ+τ))|2

)1/4.

I If U = I then this map is a Heisenberg translation.Action on N2n−1 and Cygan translation length are as before.

General parabolic maps in PSp(n, 1)

Cao-JRP 2014A general parabolic map in PSp(n, 1) has the form

T =

µ −√

2τ∗µ (−‖τ‖2 + t)µ

0 U√

2τµ0 0 µ

where U ∈ Sp(n − 1) and µ ∈ H, |µ| = 1

with

Uτ = µτ, U∗τ = µτ, µτ 6= τµ if τ 6= 0 and µ 6= ±1,

Uτ = τ, U∗τ = τ if τ 6= 0 and µ = ±1,

µt 6= tµ if τ = 0 and µ 6= ±1,

t 6= 0 if τ = 0 and µ = ±1.

This acts on N4n−1 asT : (ζ, v) 7−→ (Uζµ+ τ, µvµ+ t + 2=(µζ∗µτ))

Its Cygan translation length `T ((ζ, v)) is(‖Uζµ− ζ + τ‖4 + |µvµ− v + t + 2=((ζ∗ − τ∗)(Uζµ+ τ))|2

)1/4.

Vertical projection

Before discussing the generalised Shimizu’s lemma, there is onemore ingredient.

I Define vertical projectionΠ : N2n−1 = Cn−1 × R −→ Cn−1

Π : N4n−1 = Hn−1 × R3 −→ Hn−1

by Π : (ζ, v) 7−→ ζ.

I If T is one of the parabolic maps defined above, its verticalprojection acts on Cn−1 or Hn−1 respectively asTΠ : ζ 7−→ Uζµ+ τ (where µ = 1 in the complex case)

I The Euclidean translation length of the vertical projection ofT at ζ ∈ Fn−1 is `Π

T (ζ) = ‖Uζµ− ζ + τ‖

The generalised Shimizu’s lemmaLet T be a general parabolic map, U, µ as before.Define NU,µ = max{‖Uζµ− ζ‖ : ‖ζ‖ = 1}Nµ = max{‖µζµ− ζ‖ : ‖ζ‖ = 1} = |=(µ)|

Theorem (Cao-JRP 2014)

Let Γ ∈ PSp(n, 1) be discrete and contain parabolic T as above.Suppose Nµ < 1/4 and NU,µ < (3− 2

√2 + Nµ)/2

Define K = 12

(1 + 2NU,µ +

√1− 12NU,µ + 4N2

U,µ − 4Nµ)

Let S ∈ Γ not fixing ∞ have isometric sphere of radius rS . Then

r2S ≤

`T (S−1(∞))`T (S(∞))

K+

4`ΠT (ΠS−1(∞))`Π

T (ΠS(∞))

K (K − 2NU,µ)

I When µ = 1 (including case of PU(n, 1)) hypotheses simplify:NU = NU,1 < (

√2− 1)2/2 and

K = 12

(1 + 2NU +

√1− 12NU + 4N2

U

)The conclusion remains the same.

I When U = I , µ = 1 get version for Heisenberg translations.

Sketch of the proof when U 6= I (so NU,µ 6= 0)

I Consider the sequence S0 = S , Sj+1 = SjTS−1j

I Show in finite basin of attraction of dynamical system

I Deduce Sj tends to T .

The variables in the dynamical system are:Xj =

(max{`T (S−1

j (∞)), `T (Sj(∞))}/rSj)2,

Yj =(max{`Π

T (ΠS−1j (∞)), `Π

T (ΠSj(∞))}/rSj)2

They satisfy recursion inequalities:Xj+1 ≤ X 2

j + 4Yj + 2NU,µ + Nµ, Yj+1 ≤ XjYj + 2NU,µYj + N2U,µ

If the bound on rS in the theorem is not true then:

I Xj + 4Yj/(K − 2NU,µ) < K

I for large enough j we haveXj < K − 2NU,µ and Yj < (K − 2NU,µ)NU,µ/2

I for all ε > 0 there exists Jε so for all j ≥ Jε:Xj < 1− K + ε and Yj < (1− K )NU,µ/2 + ε

Where do we go from here?

I For screw parabolic maps T where U has infinite order,the asymptotic growth (in terms of distance from axis)of bounds on rS are worse than in examples.

I Erlandsson & Zakeri: In PO(4, 1) use same bounds forcarefully chosen powers of T to improve asymptotics.‘Carefully chosen’ means use Diophantine approximation ofrotation angle.

I Erlandsson-Zakeri’s idea also works in PU(2, 1).

I Try to generalise these results to F4(−20)

parabolic isometries of hyperbolic spaces and...

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